I understand the problem and the solution using the Hardy Cross Method except for a negative sign in the solution.
correction formula = -sigma k*Q / n * sigma absolute of k *Q
In the solution it is assumed all flows in the clockwise direction are positive.
For the numerator for the correction formula, the solution reads -{(63*1.25^2)-(382*0.72^2)-*(35*0.75^2)}. The whole sum of k*Q should be taken against the negative sign but shouldn't the first flow value of 1.25 cfs be taken as a negative as well. The flows for the other two pipe branches of 0.75 cfs are both consider negative. I don't think I explained that well but any help would be appreciated. You will have to look at the book for an illustration of the pipe network.
correction formula = -sigma k*Q / n * sigma absolute of k *Q
In the solution it is assumed all flows in the clockwise direction are positive.
For the numerator for the correction formula, the solution reads -{(63*1.25^2)-(382*0.72^2)-*(35*0.75^2)}. The whole sum of k*Q should be taken against the negative sign but shouldn't the first flow value of 1.25 cfs be taken as a negative as well. The flows for the other two pipe branches of 0.75 cfs are both consider negative. I don't think I explained that well but any help would be appreciated. You will have to look at the book for an illustration of the pipe network.
